General formula for $\sum_{n=0}^\infty \frac{1}{(an+b)^c}, a> 0, b>0, c>1$

Question

Is there a general formula for this sum? $$I(a,b,c)=\sum_{n=0}^\infty \frac{1}{(an+b)^c}, a> 0, b>0, c>1$$ I noticed that the first few a, b, and c values yielded seemingly different results, but all look related in some way to the zeta function. Here are a few values of I. $$I(2, 1, 2)= \frac{\pi^2}{8}$$ $$I(2, 1, 3)= \frac{7\zeta(3)}{8}$$ $$I(3, 1, 2)= \frac{\psi^{(1)}(\frac{1}{3})}{9}$$ $$I(1,2,2)= \frac{\pi^2}{6} - 1$$ $$I(\pi,\pi,\pi)= \frac{\zeta(\pi)}{\pi^{\pi}}$$ Don't pay too much attention to the bounds for a and b as I don't really know if anything changes should the values be negative. Perhaps if a general formula does not exist, maybe one does when fixing certain values such as setting b = 1?

Answer 1

$I(a,b,c) =\sum_{n=0}^\infty \frac{1}{(an+b)^c} $

Some simple cases.

$I(1,1,c) =\sum_{n=0}^\infty \frac{1}{(n+1)^c} =\sum_{n=1}^\infty \frac{1}{n^c} =\zeta(c) $

$I(2,2,c) =\sum_{n=0}^\infty \frac{1}{(2n+2)^c} =2^{-c}\sum_{n=1}^\infty \frac{1}{n^c} =2^{-c}\zeta(c) $

$\begin{array}\\ I(2,1,c) &=\sum_{n=0}^\infty \frac{1}{(2n+1)^c}\\ &=\sum_{n=0}^\infty \frac{1}{(2n+1)^c}+\sum_{n=0}^\infty \frac{1}{(2n+2)^c}-\sum_{n=0}^\infty \frac{1}{(2n+2)^c}\\ &=\sum_{n=1}^\infty \frac{1}{n^c}-I(2, 2, c)\\ &=I(1, 1, c)-I(2, 2, c)\\ &=\zeta(c)-2^{-c}\zeta(c)\\ &=(1-2^{-c})\zeta(c)\\ \end{array} $

If $b$ is an integer,

$I(1,b,c) =\sum_{n=0}^\infty \frac{1}{(n+b)^c} =\sum_{n=b}^\infty \frac{1}{n^c} =\sum_{n=1}^\infty \frac{1}{n^c}-\sum_{n=1}^{b-1} \frac{1}{n^c} =\zeta(c)-\sum_{n=1}^{b-1} \frac{1}{n^c} $

Note that $I(a,b,c) =\sum_{n=0}^\infty \frac{1}{(an+b)^c} =a^{-c}\sum_{n=0}^\infty \frac{1}{(n+b/a)^c} =a^{-c}\zeta(c, b/a) $ the Hurwitz zeta function.

See https://en.wikipedia.org/wiki/Hurwitz_zeta_function for a discussion of this.

Answer 2

For $a=b=c=k$, where k is a constant, it seems as though $I(k, k, k) = \frac{\zeta(k)}{k^k}$

Answer 3

While not a complete answer, if $c$ is a positive integer greater than 1 the sum can be written in terms of the polygamma function $\psi^{(m)}(z)$.

From the series representation for the polygamma function, namely $$\psi^{(m)}(z) = (-1)^{m + 1} \Gamma (m + 1) \sum_{n = 0}^\infty \frac{1}{(n + z)^{m + 1}}, \quad m > 0, z \notin 0,\mathbb{Z}^-$$ if we rewrite your sum as $$I(a,b,c) = \frac{1}{a^c} \sum_{n = 0}^\infty \frac{1}{(n + b/a)^{(c - 1) + 1}},$$ in terms of the polygamma function the sum becomes $$I(a,b,c) = \frac{(-1)^c}{a^c \Gamma (c)} \psi^{(c - 1)} \left (\frac{b}{a} \right ).$$ Here $c > 1$ where $c \in \mathbb{N}$, $a > 0$, and $b \neq 0,-a,-2a,\ldots$